If P( 2n+1, n-1): P(2n-1,n) = 3:5, find n

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If P( 2n+1, n-1): P(2n-1,n) = 3:5, find n

by Naruto » Sun Jun 14, 2009 4:25 am
If P( 2n+1, n-1): P(2n-1,n) = 3:5, find n

A 2
B 4
C 6
D 8
E 10



Explain the question first and then the answer. :)

OA is B

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by tohellandback » Sun Jun 14, 2009 6:16 am
n=4
P( 2n+1, n-1): P(2n-1,n) = 3:5
or, (2n+1)!/(n+2)!: (2n-1)!/(n-1)!=3/5 (using the permutations formula npr=n!/(n-r)!

or(2n+1)!/(n+2)! * (n-1)!/(2n-1)!=3/5

2n(2n+1)/n(n+1)(n+2)=3/5

solve this..you will get n=4. reject he other value i.e n=-1/3
The powers of two are bloody impolite!!